= Relativistic radio-cavity enthalpy
{title2=$H_{\rm cav}=\gamma pV/(\gamma-1)\simeq4pV$}
For an approximately pressure-balanced cavity containing relativistic fluid with $\gamma=4/3$, internal <energy> plus displacement work is $4pV$. Dividing by an estimated inflation or buoyancy age gives a mechanical-power diagnostic. Projection, pressure <gradients>, the equation of state, age estimates and additional <shock> <energy> affect the inference; cavity power is not identical to locally dissipated heat.
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